Answered

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a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem to find the girl's initial velocity when she began to stop?

Sagot :

W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j 
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

Answer:

10 m/s

Explanation:

The work-kinetic energy theorem states that the net work done on an object is equal to the change in the kinetic energy of the object.

In formula:

[tex]W=K_f -K_i[/tex] (1)

where

W is the work done

Ki is the initial kinetic energy

Kf is the final kinetic energy

In this problem, we have:

[tex]W=-5000 J[/tex] the net work done on the gymnast

[tex]m=100 kg[/tex] is the mass of the gymnast

[tex]v_f = 0[/tex] is the final velocity of the gymnast, so her final kinetic energy is also zero:

[tex]K_f = \frac{1}{2}mv_f^2 = 0[/tex]

Therefore, we can rewrite eq.(1) as

[tex]W=-\frac{1}{2}mv_i^2[/tex]

where [tex]v_i[/tex] is the initial velocity of the girl. By substituting the numbers and re-arranging the equation, we find:

[tex]v_i = \sqrt{-\frac{2W}{m}}=\sqrt{-\frac{2(-5000 J)}{100 kg}}=10 m/s[/tex]

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